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Question:

The numbers −16, 2^x+3 − 2^2x−1 − 16, 2^2x−1 + 16 are in an arithmetic progression. Then x equals ________.

Explanation:

−16, 2^x+3 − 2^2x−1 − 16, 2^2x−1 + 16 are in AP

If a, b and c are in AP ⇒ 2b = a + c

∴ 2 × (2^x+3 − 2^2x−1 − 16) = -16 + 2^2x−1 + 16

⇒ 2^x+4 − 2^2x − 32) = 2^2x−1

⇒ 16 × 2^x − 2^2x − 32 = 2^2x​​/2

⇒ 32 × 2^x − 2 × 2^2x − 64 = 2^2x​​

Take 2^x = a and 2^2x = a² 

⇒ 32a − 2a² − 64 = a²​​

⇒ 3a²​​ - 32a + 64 = 0

⇒ 3a²​​ - 8a - 24a + 64 = 0

⇒ (3a - 8)(a - 8) = 0

⇒ a = 8/3 or 8

⇒ 2^x = 8/3 or 8 

[8/3 is rejected as 2^x cannot be a fraction]

 

⇒ 2^x = 8 = 2³ 

⇒ x = 3 

Hence, 3.